The function $f ( x ) = ( 2 x - 1 ) ^ { 2 }$ is one-to-one if its domain is

A) $( - \infty , \infty )$
B) $\left( - \infty , \frac { 1 } { 2 } \right)$
C) $[ - 1 , \infty )$
D) $( - \infty , - 1 ) \cup ( 1 , \infty )$
E) $( - \infty , 2 ]$

Question

The function $f ( x ) = ( 2 x - 1 ) ^ { 2 }$ is one-to-one if its domain is

A) $( - \infty , \infty )$
B) $\left( - \infty , \frac { 1 } { 2 } \right)$
C) $[ - 1 , \infty )$
D) $( - \infty , - 1 ) \cup ( 1 , \infty )$
E) $( - \infty , 2 ]$

Quizplus · Accepted Answer

The function $f(x) = (2x - 1)^2$ is a quadratic function and is not one-to-one over its entire domain because it is a parabola opening upwards. However, if we restrict the domain so that the function is either strictly increasing or decreasing, it can be one-to-one. The vertex of this parabola is at $x = \frac{1}{2}$, and to the left of this point, the function is strictly decreasing. Therefore, restricting the domain to $\left( - \infty, \frac{1}{2} \right)$ makes the function one-to-one.

The Function f(x)=(2x−1)2f ( x ) = ( 2 x - 1 ) ^ { 2 }f(x)=(2x−1)2

The Function $f ( x ) = ( 2 x - 1 ) ^ { 2 }$